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Different sinewaves to each end of single wire

Well, here's how I'd think things would work out, in the sort of setup that your (not entirely clear) description would seem to indicate.

If I understand correctly: you have two independent voltage generators (e.g. audio amplifier channnels). Each one's "hot" output is connected through a series resistor of some sort, to a common load point (this is often called a "build-out" resistor). The two generators' "ground" outputs are connected together to the other side of the load point. The two amps/generators are assumed to have independent power supplies, with no common reference other than the two connections at the inputs.

Let's assume, to simply matters, that there is no load at the common load point, other than what the two voltage generators themselves might create for one another.

Let's also assume that each generator is "perfect", in that it has a zero-ohm output impedance (other than what's provided by its series build-out resistor) and can provide unlimited current.

Let's also assume that the two build-out resistors are identical in value.

Let's look at several cases:

- The two generators are operating at the same frequency, the same amplitude, and are exactly in phase with one another. In this case, the voltage at the common load point is going to always be exactly the voltage of each generator's output. The chain from "generator 1 output, through generator 1 build-out resistor, to common load point, through generator 2 build-out resistor, to generator 2 output" will be all at the same voltage, at any given instant in time, as there's nothing anywhere "pulling" it away from this voltage. In effect, each generator will see a "no real load" situation. No current will flow.

- One generator is operating as above. The other is set to "0 Hz, 0 amplitude, 0 degrees phase angle" - it's always "generating" zero volts. In effect, the second generator's output is a perfect short to ground. In this case, generator 1 will see a load consisting of the two build-out resistors in series (i.e. the sum of the two). It will deliver an AC current into these resistors which will be (at any given instant) the instantaneous voltage it's delivering, divided by the sum of the two resistor values. Generator 2 will be delivering exactly the same current, but 180 degrees out of phase. It will be "seeing" something that's not a simple resistive load, as the current drawn by the "load" (generator 1) is not going to be proportional to the voltage that this generator is trying to deliver.

- One generator is operating as in the first case. The second generator is generating the same frequency and amplitude, but 180 degrees out of phase with the first. In this case, the voltage at the common load point will always be zero... the two generators are delivering identical currents to this point, through identical resistors - they're "tugging" equally hard but in opposite directions. This is often referred to as a "virtual ground". From the point of view of each generator, it's loaded into a perfect short circuit to ground. It will deliver an current equal at any given instant to the voltage it's trying to deliver (at that point in the sinewave) divided by the value of that generator's build-out resistor. Once again, the two generators will be delivering exactly the same current, but 180 degrees out of phase with one another.

- The two generators are generating different frequencies.

In this case, the current flow out of generator 1 is going to be equal, at any given instant, to the differences in output voltages of the two generators, divided by the sum of the two build-out resistors. If you plot this, you'll find that the current is a "difference of sines". The current flow out of generator 2 will be exactly the same, but of opposite polarity.

This final case is actually the "general case" for this problem, I believe. The first three cases are simply special cases of this. In every case where current flows (i.e. other than the first) you'll be radiating at least a small amount of energy away as EM.

--
Dave Platt                                    AE6EO 
Friends of Jade Warrior home page:  http://www.radagast.org/jade-warrior 
  I do _not_ wish to receive unsolicited commercial email, and I will 
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Reply to
Dave Platt
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In a lossless system, then, yes, nothing tangible is happening. For the balls banging into one another, there is loss. The air is heated by friction, there's sound emitted, the collisions are inelastic, so the balls heat up as the amplitude decreases toward zero. If you add it all up, the potential energy of the ball when you let it go is converted to exactly that amount of those other things.

I don't think I said that the net work is zero. I said the net electron flow is zero. If you stuff current into a speaker, the cone deflects and produces sound. But no electrons were consumed in the process. You just moved them from one side to the other. On the opposite half cycle, you moved them back. Net electron flow is still zero. It takes energy to push those electrons. The net energy required is positive no matter which directions the electrons are flowing at any instant. For a resistive circuit, Instantaneous power P=I^2*R. The square takes care of the sign.

The answer to your original question is simple. NO! You can't cancel one frequency with a different frequency. The interweb is not known for patience or virtue.

Reply to
mike

Many thanks for your detailed reply. This seems to be most like the answer I was looking for.

The current moves back and forth within the conductor/coil, summing to zero, and an EMF is thereby generated at the sinewaves' difference frequency.

As previously mentioned the two driving amps would have the same internal ground reference.

Original question is below for reference.

Kevin Summers

*************

What happens when you feed a sinewave into each end of a length of straight wire, but the sinewaves differ in frequency. Amplitudes are the same. There is no connected ground reference.

Is there an internal current flow between the higher and lower frequency ends of the wire? If so, how can it calculated or measured?

Reply to
ksummers

Oops! Sorry, I didn't catch this prior to replying earlier.

Your above description does not actually reflect my OP. The only ground is that within the stereo amp itself.

Each "hot" terminal of the amp is connected a _different_ end of the conductor/coil. They are not wired together at one end, and there is no external ground connection.

What I am hoping to demonstrate under the above condition is that, even though there is no net current flow, an EMF is emitted, at the difference frequency of the two signals.

Kevin Summers

Reply to
ksummers

IN a linear system, there is no difference frequency. There is one waveform that can be modeled by the superposition of two sinewaves.

Reply to
mike

OK, gotcha. I think I see the picture...

... and I disagree with your suggested analysis.

Your system analyzes out very close to what I had proposed. Instead of a single "common load point" between the two amp outputs, there's a length of wire.

Most of the analysis I suggested, still applies. As long as the length of the wire is fairly short, compared to the electrical wavelengths of the frequencies in question, it will behave a lot like a short circuit between these two points (or a small-valued resistor).

What's going to happen in this case, is that there will be electrical currents flowing back and forth through the wire, in every case but one (and that special case is the one where the two amp channels are exactly identical, and the wire is very short).

Each amp channel is generating a "pure AC" signal (in this case)... that is, there's no DC voltage, just a pure alternating voltage at the frequency in question.

So, there will be no net DC (!) current flow from either amplifier. However, each amplifier will be outputting an AC current, which will generate an AC electromagnetic field around the wire as it flows through the wire towards the other end.

The frequency of the EM field created around the wire by amp channel

1, will be at the frequency that amp channel 1 is creating. Similarly for amp channel 2.

If the two amplifiers are "perfect" ones (i.e. they don't distort, or clip, or otherwise exhibit any form of nonlinear behavior) then these are the *only* two frequencies you'll see in the EM field.

For example, if channel 1 is at 100 Hz, and channel 2 is at 110 Hz, then the EM field you measure will have 100 Hz and 110 Hz components, and that's it. You will *not* measure any signal energy at 10 Hz (the difference) or 210 Hz (the sum).

The funny thing is that if you look at the measured waveform, it will

*appear* to pulse at a 10 Hz rate - the "beat frequency". However, there isn't actually any energy at that frequency, if the whole generation and measurement process is linear.

If any non-linearity creeps into the electronics or measurement system along the way, though, then both the sum (210 Hz) and difference (10 Hz) frequencies can actually be created... this is the process known as "mixing" (which is different, mathematically, than the plain "addition" process you get in a purely linear system).

--
Dave Platt                                    AE6EO 
Friends of Jade Warrior home page:  http://www.radagast.org/jade-warrior 
  I do _not_ wish to receive unsolicited commercial email, and I will 
     boycott any company which has the gall to send me such ads!
Reply to
Dave Platt

I think I may have put that in your head. Let me try to undo it. Think of electrons as cars. Stand on the corner and count cars. In the morning, people go to work and your car count goes up. In the evening, all those people go back home in the opposite direction and your car count goes back down to zero. The definition of current is the number of electrons per second passing a point in space in the same direction. The instantaneous traffic is not zero...it's positive in the AM and negative in the PM. But at midnight, when all the cars are back home in the garage, the net transfer averaged over a day cycle is zero. If you have multiple lanes and cars going in both directions, the sum of the counts is equal to the count of the sum. They add, but they don't interact.

Your basic problem is that you refuse to let go of the concept of difference frequency. In a linear, time-invariant system, there is no difference frequency. All the wishing in the world won't change that. You have EMF generated by one signal. You have EMF generated by the second signal. And the sum of the EMFs is exactly the same as the EMF generated by the sum of the signals.

A*sin(F1) + B*sin(F2) = signal THERE IS NO DIFFERENCE FREQUENCY, ANYWHERE, NONE, NOT ANY.

Reply to
mike

what does 'feed' mean?

Pretty poorly stated question.

Reply to
Robert Macy

Yes, now I understand ... it's linear. Not to beat a dead horse, but what if the aforementioned wire/coil, driven at each end by a dissimilar frequency, was wrapped around a ferrite core?

The latter would enable non-linear mixing, and perhaps a real difference frequency. Time to try this out on the bench.

Many thanks for your explanations.

Kevin Summers

Reply to
ksummers

Hmm... I think this is what threw me off. Apart from not being a math genius.

I will make the "wire" into a coil wrapped around ferrite and see what happens then on the CRO.

Many thanks for your help.

Kevin Summers

Reply to
ksummers

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